Orientation · First Class

Welcome to Calculus

A new kind of calculation — two operations, one chain of ideas, and the single question that starts it all.

1. Calculus is a kind of calculation

You have been calculating for years. In arithmetic you learned four operations that act on numbers: addition, subtraction, multiplication and division. Give the operation two numbers, it hands you back a number. In algebra you learned rules for those operations — a distributive law, an order of operations, shortcuts for powers — and most of your work was applying those rules correctly.

Calculus is the same idea, one level up. It gives you two brand-new operations, and they do not act on numbers. They act on whole functions, and they hand you back another function.

⚡ Differentiation

Input a function $f$. Output a new function $f'$ that reports how fast $f$ is changing at every point.

$$f(x)=x^2 \;\longmapsto\; f'(x)=2x$$

📦 Integration

Input a function $f$. Output a new function that reports how much of $f$ has accumulated so far.

$$f(x)=2x \;\longmapsto\; \int_0^x 2t\,dt = x^2$$

Look carefully at those two boxes. They are the same pair of functions, read in opposite directions. That is not a coincidence, and by the end of this course it will be the single most important fact you know.

🧭 The plan for this course

Almost everything ahead falls into one of three jobs:

Learn the two operations — what $\dfrac{d}{dx}$ and $\displaystyle\int$ mean, and the rules for computing them quickly (power rule, product rule, chain rule, substitution).
Use them — rates of change, maxima and minima, optimization, areas, volumes, accumulated totals.
Understand why they work — which is where limits and continuity come in, and why we start there.

🗺️ Course roadmap — why we start with limits

Differentiation and integration are both defined by limits, so the course builds the limit first and every later chapter stands on it:

Continuity (1.3) — $f$ is continuous at $a$ exactly when $\displaystyle\lim_{x\to a}f(x)=f(a)$.
The derivative (2.1) — $f'(x)=\displaystyle\lim_{h\to0}\frac{f(x+h)-f(x)}{h}$. Putting $h=0$ gives $\tfrac00$, so only the limit works; the algebra of 1.2 (factor, cancel, rationalise) is exactly what finding derivatives needs.
The definite integral (4.2) — area is the limit of sums of thin rectangles, $\displaystyle\int_a^b f(x)\,dx=\lim_{n\to\infty}\sum_{i=1}^{n}f(x_i^*)\,\Delta x$. This limit has $n\to\infty$, which is why limits involving infinity (1.4) are in Chapter 1; they are also used for asymptotes and L'Hospital's rule.

Pre-calculus → Limits → Continuity → Derivative → Rules → Applications (Ch. 3)
Limits at infinity → Definite integral ⟵ FTC ⟵ Antiderivatives (reverse of the derivative)

Chapter by chapter. Chapter 1 finds limits from graphs, tables and algebra and uses them to define continuity; continuity comes before the derivative because a function must be continuous to be differentiable (though a corner can be continuous and still not differentiable). Chapter 2 defines the derivative as a limit, then proves the rules once (power, product, quotient, chain, implicit, parametric) so we can use them as shortcuts. Chapter 3 applies derivatives: tangent and normal lines, motion, related rates, maxima and minima, the Mean Value Theorem, optimization, and L'Hospital's rule, which closes the loop by using derivatives to evaluate hard limits. Chapter 4 reverses differentiation, defines the definite integral as accumulated area, and joins the two halves of the course through the Fundamental Theorem, $\displaystyle\int_a^b f(x)\,dx=F(b)-F(a)$.

Threads to watch for: every new definition is a limit in disguise; rules are shortcuts, not new ideas; each idea is shown as a graph, a table and a formula; rate and accumulation are opposite operations.

2. The two operations, side by side

Before we do any theory, watch the two operations at work on the same function. On the left, a point slides along the curve and the machine reports the slope there — that is differentiation. On the right, the same slide fills in the area collected so far — that is integration.

Keep one eye on the bottom readout. It compares the height of the curve with the growth rate of the shaded area.

Function: Position c:

⚡ Derivative — the slope at c

📦 Integral — the area up to c

position $c$0

height $f(c)$0

slope $f'(c)$0

area $A(c)$0

🔗 What you just saw — the Fundamental Theorem, in advance

The rate at which the shaded area grows is exactly the height of the curve: $$\frac{d}{dx}\int_a^x f(t)\,dt = f(x).$$ Differentiation and integration undo each other. We will prove this in Chapter 4, but it is worth meeting on day one, because it tells you the whole course is about one idea seen from two sides, not two unrelated techniques.

3. So let us try to differentiate — and watch it fail

Here is where the course really begins. Differentiation asks for the slope of a curve at a single point. But a slope needs two points. The trick is to take a second point nearby, at distance $h$, draw the straight line through the two (a secant line), and then slide the second point closer and closer.

$$\text{slope of the secant} \;=\; \frac{f(a+h)-f(a)}{h}$$

If those secant slopes settle down to one number as $h$ shrinks, we call that number the derivative. The interactive below runs exactly that experiment on five different curves. One of them works. Four of them break — and how they break is the reason limits and continuity exist.

⚠️ The lesson from the broken cases

The jump and the hole failed for the same underlying reason: near $x=a$ the function's values did not settle down to $f(a)$. One of them leapt to a different level; the other had no value at $a$ at all.

So before we can ask for a derivative, we need a name for "the graph is unbroken here". That name is continuity.

4. What continuity actually says

Informally, $f$ is continuous at $a$ if you can draw the graph through $x=a$ without lifting your pen. To make that precise we need to compare two things: where the function is heading as you approach $a$, and where the function actually is at $a$.

📘 Definition — Continuity at a point

A function $f$ is continuous at $a$ if all three of these hold:

$f(a)$ is defined,
$\displaystyle\lim_{x\to a}f(x)$ exists,
$\displaystyle\lim_{x\to a}f(x)=f(a)$.

If any one fails, $f$ is discontinuous at $a$. We study this properly in Section 1.3.

Read condition 2 again. The definition of continuity contains a limit. We cannot define continuity until we can say what $\displaystyle\lim_{x\to a}f(x)$ means. And so the chain pulls us back one more step.

✅ Continuous at $a$ — the limit and the value agree.

❌ No value at $a$ — condition 1 fails.

❌ The value sits off the curve — condition 3 fails.

5. The bottom of the chain: the limit

A limit answers one question, and it is a strangely careful question:

📘 The limit question

As $x$ gets closer and closer to $a$ — without ever being $a$ — what single number do the values $f(x)$ get close to and stay close to?

If that number is $L$, we write $\displaystyle\lim_{x\to a}f(x)=L$.

Two things about that question are deliberate and easy to miss.

We never let $x$ equal $a$. That is exactly why the limit can rescue the hole case above: $\dfrac{x^2-1}{x-1}$ has no value at $x=1$, yet everywhere near $1$ it equals $x+1$, so it heads confidently to $2$. The limit sees where a function is going even when the function never gets there.

The answer must be the same from both sides. That is why the jump case has no limit at $x=1$: the left side heads to $1$, the right side heads to $3$, so there is no single destination.

That is the whole idea, and Section 1.1 is devoted to it — reading limits from tables, reading them from graphs, and recognising when they fail to exist.

6. Now the chain runs forward

We arrived here by walking backwards, each step forced by the one before it. Run the same chain forwards and it becomes the syllabus.

🧩 Preview — the two places limits are hiding

Both new operations are defined as limits. This is why Chapter 1 has to come first. $$f'(a)=\lim_{h\to 0}\frac{f(a+h)-f(a)}{h} \qquad\qquad \int_a^b f(x)\,dx=\lim_{n\to\infty}\sum_{i=1}^{n} f(x_i)\,\Delta x$$ The first is a limit of slopes of secant lines. The second is a limit of totals of thin rectangles. Same machinery, two different questions.

7. Three examples to end the first class

✏️ Example 1 — A limit turning into a derivative

A ball falls so that its distance after $t$ seconds is $s(t)=t^2$ metres. Find its instantaneous speed at $t=1$.

💡 Show solution▶

Step 1 Speed over an interval is a ratio we already know: distance divided by time. From $t=1$ to $t=1+h$, $$\text{average speed}=\frac{s(1+h)-s(1)}{h}=\frac{(1+h)^2-1}{h}.$$

Step 2 Expand and simplify, remembering that $h\neq 0$ so we may cancel: $$\frac{1+2h+h^2-1}{h}=\frac{2h+h^2}{h}=2+h.$$

Step 3 Now let the interval shrink. As $h\to 0$ the average speeds $2+h$ head to $2$: $$s'(1)=\lim_{h\to 0}(2+h)=2\ \text{m/s}.$$

Note We could not simply set $h=0$ at the start — that gives $\tfrac00$. The limit is what lets us take $h$ as close to zero as we like without ever landing on it. That single move is the whole of Chapter 2.

✏️ Example 2 — Continuous, yet no derivative

Let $f(x)=|x|$. Show that $f$ is continuous at $x=0$ but has no derivative there.

💡 Show solution▶

Step 1 Continuity. Approaching $0$ from the left, $f(x)=-x\to 0$; from the right, $f(x)=x\to 0$. So $\displaystyle\lim_{x\to0}|x|=0$, and $f(0)=0$. All three conditions hold, so $f$ is continuous at $0$.

Step 2 The difference quotient. For $h>0$, $\dfrac{|0+h|-|0|}{h}=\dfrac{h}{h}=1$. For $h<0$, $\dfrac{-h}{h}=-1$.

Step 3 The two sides head to different numbers, $1$ and $-1$, so $\displaystyle\lim_{h\to0}\frac{|h|}{h}$ does not exist and $f'(0)$ does not exist.

Moral Continuity is necessary for differentiability but not sufficient. Unbroken is not enough; the curve must also have one single direction at the point. Try the "Corner" option in Interactive 2 to watch this happen live.

✏️ Example 3 — Accumulation, and the loop closing

Water flows into a tank at the rate $r(t)=2t$ litres per minute. How much water enters during the first $3$ minutes, and what is the connection to differentiation?

💡 Show solution▶

Step 1 The rate is not constant, so "rate × time" is not allowed directly. Chop the three minutes into thin slices. On a slice so short that $r$ barely changes, the water added is about $r(t)\,\Delta t$ — the area of a thin rectangle under the rate graph.

Step 2 Add the slices and let them get thinner. The total is the area under $y=2t$ from $0$ to $3$, a triangle of base $3$ and height $6$: $$V=\int_0^3 2t\,dt=\tfrac12(3)(6)=9\ \text{litres}.$$

Step 3 Now let the end time be a variable: $V(x)=\displaystyle\int_0^x 2t\,dt = x^2$. Differentiate it: $V'(x)=2x=r(x)$.

Moral The accumulated volume differentiates back into the rate we started with. Accumulating and rate-finding are inverse operations — the Fundamental Theorem, met here as an observation and proved in Chapter 4.

Quick checks

Six short prompts to close the class. Try each one before opening the solution.

For $s(t)=t^2$, watch the average speeds over $[1,\,1+h]$ as $h$ shrinks. What number are they heading for, and why can we not simply put $h=0$?

Average speeds as $h$ shrinks

$h$	interval	average speed

💡 Show solution▶

Step 1 The averages are $2.5,\ 2.1,\ 2.01,\ 2.001,\dots$ — heading to $2$.

Step 2 Setting $h=0$ gives $\dfrac{0}{0}$, which is meaningless. The expression simplifies to $2+h$ only when $h\neq0$, and the limit lets us use values as near $0$ as we please without ever using $0$ itself. So $s'(1)=2$ m/s.

At which of the three marked places on the graph below does the function fail to have a derivative? Give the reason for each.

Where does differentiation break?

💡 Show solution▶

$x=-1$ A corner. The function is continuous, but the slope coming in from the left ($-1$) differs from the slope going out to the right ($+1$), so there is no single tangent direction.

$x=1$ A hole. $f(1)$ is not defined, so there is no point on the curve to anchor a tangent to. The limit still exists here, which is why this break is called removable.

$x=3$ A jump. The left and right sides head to different heights, so the limit fails, continuity fails, and the derivative fails with it.

Everywhere else The curve is smooth and unbroken, so the derivative exists.

Water enters a tank at $r(t)=2t$ L/min. Use the animation to estimate the volume after $3$ minutes by adding thin rectangles, then compare with the exact answer.

Rectangles filling the tank

Rectangles:

number of rectangles4

rectangle total (litres)0

exact area (litres)9

💡 Show solution▶

Step 1 These are left-endpoint rectangles, so each one sits a little under the curve and the total is an underestimate. With four rectangles it is $6.75$; with forty it is $8.775$, and it keeps climbing toward $9$.

Step 2 Exactly: the region is a triangle of base $3$ and height $r(3)=6$, so $\displaystyle\int_0^3 2t\,dt=\tfrac12(3)(6)=9$ litres.

Step 3 Notice the structure — a sum of many small pieces, then a limit as the pieces get small. That is the definition of the definite integral, and it is the same limiting move that produced the derivative.

Sort these questions into "needs a derivative" and "needs an integral": (a) How fast is the balloon rising right now? (b) How much fuel did the engine burn over the whole trip? (c) Which box design uses the least cardboard? (d) What is the area between two curves?

💡 Show solution▶

Derivative (a) "Right now" means an instantaneous rate. (c) Least or most means a maximum or minimum, which we locate by asking where the rate of change is zero — that is Chapter 3.

Integral (b) A total collected over a whole interval from a varying rate. (d) Area is an accumulation of thin strips.

Rule of thumb "How fast, how steep, when is it biggest" $\Rightarrow$ derivative. "How much in total, how far, what area" $\Rightarrow$ integral.

Put these four ideas in the order a strict logical development must introduce them, and say why each is needed: derivative, continuity, limit, integral.

💡 Show solution▶

Order limit $\to$ continuity $\to$ derivative $\to$ integral.

Why Continuity is defined using a limit. The derivative needs the function to be unbroken at the point, and is itself a limit of difference quotients. The integral is a limit of sums, and the Fundamental Theorem ties it back to the derivative, so it comes last.

But note That is the logical order, not the motivating order. We discovered the need for each one by walking backwards from the derivative, which is why this first class ran in reverse.

True or false: (a) If $f$ is continuous at $a$ then $f'(a)$ exists. (b) If $f'(a)$ exists then $f$ is continuous at $a$. (c) If $f(a)$ is undefined then $\displaystyle\lim_{x\to a}f(x)$ cannot exist.

💡 Show solution▶

False (a) $f(x)=|x|$ at $a=0$ is continuous with no derivative. Continuity is necessary, not sufficient.

True (b) Differentiability is the stronger property; it forces continuity. We prove this in Chapter 2.

False (c) $\dfrac{x^2-1}{x-1}$ is undefined at $1$, yet its limit there is $2$. A limit describes where a function is heading, not where it is.

🧪 Self-check quiz

0/5 correct

Q1. What does the derivative of a function tell you?

Q2. Why can we not simply substitute $h=0$ into $\dfrac{f(a+h)-f(a)}{h}$?

Q3. Which statement is correct?

Q4. The definition of continuity at $a$ requires $\displaystyle\lim_{x\to a}f(x)=f(a)$. What does that force us to study first?

Q5. If $\displaystyle A(x)=\int_0^x f(t)\,dt$, what is $A'(x)$?

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